1

### JEE Main 2018 (Online) 16th April Morning Slot

The relative uncertainly in the period of a satellite orbiting around the earth is 10-2. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :
A
10$-$2
B
2 $\times$ 10$-$2
C
3 $\times$ 10$-$2
D
6 $\times$ 10$-$2

## Explanation

From kepler's law,

T = 2$\pi$ $\sqrt {{{{r^3}} \over {GM}}}$

$\Rightarrow $$\,\,\,T2 = {{4{\pi ^2}} \over {GM}}{r^3} \Rightarrow$$\,\,\,$ M = ${{4{\pi ^2}} \over G} \times {{{r^3}} \over {{T^2}}}$

$\therefore\,\,\,$ ${{\Delta M} \over M}$ = 2 ${{\Delta T} \over T}$ + 3 ${{\Delta r} \over r}$

as ${{\Delta r} \over r} \simeq 0$

$\therefore\,\,\,$ $\left| {{{\Delta M} \over M}} \right|$ = 2 ${{\Delta T} \over T}$ = 2 $\times$ 10$-$2
2

### JEE Main 2019 (Online) 9th January Morning Slot

A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is :
A
2.0 %
B
2.5 %
C
1.0 %
D
0.5 %

## Explanation

We know,

$R = {{\rho l} \over A}$

and Volume (V) = A$l$

$\Rightarrow$   A $=$ ${V \over l}$

$\therefore$   $R = {{\rho {l^2}} \over v}$

$\therefore$   ${{\Delta R} \over R} = 2{{\Delta l} \over l}$

$=$   2 $\times$ 0.5

$=$ 1%
3

### JEE Main 2019 (Online) 9th January Evening Slot

Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :
A
$\sqrt {{{h{c^5}} \over G}}$
B
$\sqrt {{{{c^3}} \over {Gh}}}$
C
$\sqrt {{{Gh} \over {{c^5}}}}$
D
$\sqrt {{{Gh} \over {{c^3}}}}$

## Explanation

Let t $\propto$ Gx hy cz

$\therefore$   [t] = [G]x [h]y [c]z    . . . . . (1)

We know,

F = ${{G{M^2}} \over {{R^2}}}$

$\Rightarrow$  G = ${{F{R^2}} \over {{M^2}}}$

$\therefore$  [G] = ${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}$

[G] = $[{M^{ - 1}}{L^3}{T^{ - 2}}]$

Also,

E = hf

$\therefore$  [h] = ${{[E]} \over {[F]}}$

= ${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {[{T^{ - 1}}]}}$

= $\left[ {M{L^2}{T^{ - 1}}} \right]$

[C] = $\left[ {{M^o}L\,{T^{ - 1}}} \right]$

From equation (1) we get,

$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}$

$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]$

By comparing the power of M, L, T

$-$ x + y = 0

$\Rightarrow$  x = y

3x + 2y + z = 0

$\Rightarrow$  5x + z = 0 . . . . . (2)

$-$ 2x $-$ y $-$ z = 1

$\Rightarrow$  $-$ 3x $-$ z = 1  . . . . (3)

By solving (2) and (3), we get,

x = ${1 \over 2}$ = y and z = $-$ ${5 \over 2}$

$\therefore$  t $\propto$ $\sqrt {{{Gh} \over {{C^5}}}}$
4

### JEE Main 2019 (Online) 9th January Evening Slot

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
A
5.755 mm
B
5.950 mm
C
5.725 mm
D
5.740 mm

## Explanation

We know,

Least count (LC) = ${{Pitch} \over {no.\,of\,divisions}}$

$\therefore$  LC = ${{0.5} \over {100}}$

= 0.5 $\times$ 10$-$2 mm

Reading = MSR + CSR $-$ positive error

Given, Main scale reading (MSR) = 5.5 mm

= 48 $\times$ 0.5 $\times$ 10$-$2 mm

= 0.24

As zero of its circular scale lines 3 division below the mean line, it means error is position error.

$\therefore$  positive error

= 3 $\times$ 0.5 $\times$ 10$-$2 mm

= 0.015 mm

$\therefore$  Reading = 5.5 + 0.24 $-$ 0.015

= 5.725 mm